Question

Mam, I want the solution for these problems.1] One number is three times another.If 15 is added to the numbers, then one of the new numbers becomes twice that of the other new number. Find the numbers?2] The width of a garden 2/3 of its length. If its perimeter is 50m. Find the lenght?3] Solve: 2/3x +1/2=3/5x -5/64] The digit in the tens placeof a two digit number is three times that in ones place.If the digits are reversed, the new number will be 36 less than the original number. Find the original number

Answer 1

1] Let the numbers be x and y, so,

x = 3y......(1)

and, x + 15 = 2(y + 15)

x + 15 = 2y + 30

x - 2y = 15......(2)

Putting (1) in (2), we get

3y - 2y = 15

so, y = 15

therefore x = 3y, which gives x = 45

Hence, the numbers are 15 and 45.

2] Let width and length be W and L resp, so

W = 2/3 L.....(1)

and Perimeter is 2(L + W) which is given 50

so, 2(L + 2/3 L) = 50

L + 2/3L = 25

Solving this, (3L + 2L)/ 3 = 25

5/3 L = 25

5L = 75

giving L = 25

Therefore length is 25 m.

3] 2/3 x + 1/2 = 3/5 x - 5/6

2/3 x - 3/5 x = -5/6 - 1/2

10/15 x - 9/15 x = -5/6 -3/6

1/15 x = -8/6

Cross multiplying, we get

x = -120/6

x= -20

4] Let digit in tens place be x and ones place be y

So our number becomes 10x + y

We are given x = 3y......(1)

So, reversing digits we get 10y + x as our new number, so

10x + y - (10y + x) = 36

10x + y - 10y - x = 36

9x - 9y = 36

x - y = 4......(2)

Solving (1) and (2), we get

3y - y = 4

2y = 4

So, y = 2 and x = 6

Therefore original number is 62.